The Question is from neet-physics motion-in-a-plane resoulution-of-vectors, and it was created by the author VENKATESWARARAO

Question

$\mathop {{\rm{ }}A}\limits^ -   = \mathop {{\rm{ }}B}\limits^ -   + \mathop {{\rm{ }}C}\limits^ -  $ and the magnitudes of $\mathop {{\rm{ }}A}\limits^ -  ,\mathop {{\rm{ }}B}\limits^ -  ,\mathop {{\rm{ }}C}\limits^ -  $ are 5, 4 and 3 units respectively, the angle between $\mathop {{\rm{ }}A}\limits^ -  \mathop {{\rm{ }}C}\limits^ -  $ is:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Vector addition and dot product.
	**Related Topics**: Magnitude of a vector, angle between vectors, trigonometry
	**Prerequisites**: Basic vector algebra, dot product definition, cosine function.
## SOLUTION PATHWAY
**Given Information**:
	*  $\mathop A\limits^ -  = \mathop B\limits^ -  + \mathop C\limits^ - $
	*  $|\mathop A\limits^ -  | = 5$
	*  $|\mathop B\limits^ -  | = 4$
	*  $|\mathop C\limits^ -  | = 3$
**Method & Steps**:
	1. Isolate vector C:
	   $\mathop C\limits^ -  = \mathop A\limits^ -  - \mathop B\limits^ -  $
	   Reasoning: We need to find the angle between A and C, so we need a relation between A and C
	   Formula/Rule: Vector subtraction.
	2. Find the dot product of C with itself:
	   $\mathop C\limits^ -  \cdot \mathop C\limits^ -  = (\mathop A\limits^ -  - \mathop B\limits^ -  ) \cdot (\mathop A\limits^ -  - \mathop B\limits^ -  )$
	   Reasoning: Expanding the dot product gives us a relation with magnitudes.
	   Working: $\mathop C\limits^ -  \cdot \mathop C\limits^ -  = \mathop A\limits^ -  \cdot \mathop A\limits^ -  - 2\mathop A\limits^ -  \cdot \mathop B\limits^ -  + \mathop B\limits^ -  \cdot \mathop B\limits^ -  $
	   Key Point: Remember, $\mathop{V}\limits^ -  \cdot \mathop{V}\limits^ -  = |\mathop{V}\limits^ -  |^2$
	3. Find the dot product of A and C:
	   $\mathop A\limits^ -  \cdot \mathop C\limits^ -  = \mathop A\limits^ -  \cdot (\mathop A\limits^ -  - \mathop B\limits^ -  ) = \mathop A\limits^ -  \cdot \mathop A\limits^ -  - \mathop A\limits^ -  \cdot \mathop B\limits^ -  $
	   Reasoning: We need the dot product to find the angle
	   Formula/Rule: Vector dot product property.
	4. Find the magnitude of C:
	   $|\mathop C\limits^ -  |^2 = 3^2 = 9$
	   Reasoning: Magnitude is given.
	5. Find $\mathop A\limits^ -  \cdot \mathop B\limits^ -  $ from step 2:  $\mathop C\limits^ -  \cdot \mathop C\limits^ -  = |\mathop A\limits^ -  |^2 - 2\mathop A\limits^ -  \cdot \mathop B\limits^ -  + |\mathop B\limits^ -  |^2 \Rightarrow 9 = 25 - 2\mathop A\limits^ -  \cdot \mathop B\limits^ -  + 16 \Rightarrow \mathop A\limits^ -  \cdot \mathop B\limits^ -  = \frac{25+16-9}{2} = 16$
	6.  Substitute $\mathop A\limits^ -  \cdot \mathop B\limits^ -  $ in step 3:  $\mathop A\limits^ -  \cdot \mathop C\limits^ -  = |\mathop A\limits^ -  |^2 - \mathop A\limits^ -  \cdot \mathop B\limits^ -  = 25 - 16 = 9$
	7. Find the angle between A and C:
	   $\mathop A\limits^ -  \cdot \mathop C\limits^ -  = |\mathop A\limits^ -  ||\mathop C\limits^ -  |\cos	heta \Rightarrow 9 = 5*3*\cos	heta \Rightarrow \cos	heta = \frac{9}{15} = \frac{3}{5} \Rightarrow 	heta = {\mathop{m Cos}
olimits} ^{ - 1}\left( {\frac{3}{5}} ight)$
	   Result: The angle is ${{\mathop{m Cos}
olimits} ^{ - 1}}\left( {\frac{3}{5}} ight)$
	   Verification: The answer matches the correct option.
## OPTION ANALYSIS
[A] ✓: Correct, as derived in the solution pathway.
[B] :x: Incorrect. This would be the cosine of the angle between A and B if they formed a right triangle.
[C] :x: Incorrect. The angle is not 90 degrees. This would only be true if A and B were perpendicular, which is not the case here.
[D] :x: Incorrect, sine is not the correct trigonometric function to use here and the values are also wrong.
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly expanding the dot product, such as ignoring the cross terms or not understanding the properties of dot product.
		    [How to avoid]: Carefully apply the dot product rules: (a-b).(a-b) = a.a -2a.b +b.b, and remember that vector.vector=magnitude squared.
		2.[Error 2]: Confusing the magnitudes of vectors with the vectors themselves.
		    [How to avoid]: Always distinguish between a vector (with direction) and its magnitude (a scalar value).
	**Quick Tips**:
		1.[Helpful shortcut]: Recognizing that |A|^2 = A.A is a powerful technique to simplify dot product calculations.
		2.[Memory aid]: Remember that the dot product is also equal to the product of magnitudes and the cosine of the angle between the two vectors.

VENKATESWARARAO · Answer

The correct Option is "${{\mathop{m Cos}
olimits} ^{ - 1}}\left( {\frac{3}{5}} ight)$"

VENKATESWARARAO · Answer

${{\mathop{m Cos}
olimits} ^{ - 1}}\left( {\frac{4}{5}} ight)$

VENKATESWARARAO · Answer

$\frac{\pi }{2}$

VENKATESWARARAO · Answer

${\sin ^{ - 1}}\left( {\frac{3}{4}} \right)$

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